Câu hỏi 45

\(\mathop {\lim }\limits_{x \to  - \infty } \dfrac{{x + 2}}{{1 - 2x}} = \mathop {\lim }\limits_{x \to  - \infty } \dfrac{{1 + \dfrac{2}{x}}}{{\dfrac{1}{x} - 2}} =  - \dfrac{1}{2} \Rightarrow \left\{ \begin{array}{l}a = 1\\b = 2\end{array} \right. \Rightarrow a - b =  - 1\).

Chọn B.